Oblicz sprawność silnika Carnot'a, którego temperatura źródła ciepła wynosi 127 st.C, a chłodnica 27 st.C.
Witaj :)
dane: t₁=127⁰C, t₂=27⁰C,
szukane: η
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η = [T₁-T₂]/T₁.....T₁=273+t₁=273+127=400K......T₂=273+t₂=273+27=300K
η = [273+t₁-273-t₂]/[273+t₁] = [t₁-t₂]/[273+t₁] = [127k-27k]/[273k+127k] =
η = 100K/400K = ¼ = 0,25 = 25%
SZUKANA SPRAWNOŚĆ WYNOSI 25%.
Semper in altum....................pozdrawiam :)
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Witaj :)
dane: t₁=127⁰C, t₂=27⁰C,
szukane: η
----------------------------------
η = [T₁-T₂]/T₁.....T₁=273+t₁=273+127=400K......T₂=273+t₂=273+27=300K
η = [273+t₁-273-t₂]/[273+t₁] = [t₁-t₂]/[273+t₁] = [127k-27k]/[273k+127k] =
η = 100K/400K = ¼ = 0,25 = 25%
SZUKANA SPRAWNOŚĆ WYNOSI 25%.
Semper in altum....................pozdrawiam :)